| Iterated Function Systems Larry Riddle |
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Koch Curve
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Begin with a straight line (the blue segment in the top figure).
Divide it into three equal segments and replace the middle segment by
the two sides of an equilateral triangle of the same length as the
segment being removed (the two red segments in the middle figure).
Now repeat, taking each of the four resulting segments, dividing them
into three equal parts and replacing each of the middle segments by
two sides of an equilateral triangle (the red segments in the bottom
figure). Continue this construction.
The Koch curve is the limiting curve obtained by applying this construction an infinite number of times. For a proof that this construction does produce a "limit" that is an actual curve, i.e. the continuous image of the unit interval, see the text by Edgar. | ||||||||||
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The first iteration for the Koch curve consists of taking four copies of the original line segment, each scaled by r = 1/3. Two segments must be rotated by 60°, one counterclockwise and one clockwise. Along with the required translations, this yields the following IFS
The fixed invariant set of this IFS is the Koch curve. | ||||||||||
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Angle 60 Axiom F F --> F+F- -F+F | ||||||||||
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We have a hyperbolic IFS with each map being a similitude of ratio
r < 1. Therefore the similarity dimension, d, of the
unique invariant set of the IFS is the solution to
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Koch constructed his curve in 1904 as an example of a
non-differentiable curve, that is, a continuous curve that does not have a
tangent at any of its points. Karl Weierstrass had first
demonstrated the existence of such a curve in 1872. The article by Sime Ungar provides a simple geometric proof.
The length of the intermediate curve at the nth iteration of the construction is (4/3)^n, where n = 0 denotes the original straight line segment. Therefore the length of the Koch curve is infinite. Moreover, the length of the curve between any two points on the curve is also infinite since there is a copy of the Koch curve between any two points. Three copies of the Koch curve placed around the three sides of an equilateral triangle, form a simple closed curve that form the boundary of the Koch snowflake.
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